A uniform rod of length L = 2 m and mass M = 5 kg is placed on a table. Where is its center of mass?
The center of mass of a uniform rod lies in its center. So, the center of mass lies in 1m.
The center of mass of an object is the point where its mass is evenly distributed in all directions. It is the effective point where the entire mass of an object can be considered to act when analyzing motion and forces.
The center of mass can be calculated using:
\[ x_{CM} = \frac{m_1x_1 + m_2x_2 + m_3x_3 + \dots + m_nx_n}{m_1 + m_2 + m_3 + \dots + m_n} \]
The velocity of the center of mass of a system of particles is defined as the mass-weighted average of the individual velocities of all particles in the system. It tells you how fast and in what direction the system as a whole is moving.
Mathematically, for a system of particles:
\[\vec{v}_{\text{cm}} = \frac{1}{M} \sum_{i=1}^{n} m_i \vec{v}_i\]
If no external force acts on the system, the velocity of the center of mass remains constant due to conservation of linear momentum.
You will understand this better when you see some examples.
A uniform rod of length L = 2 m and mass M = 5 kg is placed on a table. Where is its center of mass?
The center of mass of a uniform rod lies in its center. So, the center of mass lies in 1m.
An uniform isosceles triangular plate has a base of 6 cm and a height of 8 cm. Where is its center of mass relative to the base?
\[ y_{cm} = \frac{h}{3} = \frac{8}{3} = 2.67 \text{ cm}, \quad x_{cm} = \frac{b}{2} = \frac{6}{2} = 3 \text{ cm} \]
A 5 kg object moving at 10 m/s collides with a 10 kg object at rest. After the collision, they stick together. Find their velocity after the collision using the concept of center of mass motion.
Since the two masses stick together, final velocity equals the velocity of the center of mass:
\[ V_{COM} = \frac{5 \times 10 + 10 \times 0}{5 + 10} = \frac{50}{15} = 3.33 \text{ m/s} \]
Written by Thenura Dilruk